Compound Fractions

Solving compound fractions is the important topic in algebra. Compound fractions are nothing but a mixed fraction. It is also known as the mixed number.  Compound fractions are the combination of whole number and the proper fraction. We have to write the improper fractions in the form of compound fractions. In this we have to discuss about the solving compound fractions problems.


General discussion about fractions

Definition of fraction:

Dividing two numbers is known as fractions. It takes the original form of a/b. Here we can called a as a numerator value and b as a denominator value.

Definition of proper fraction:

When the numerator value is lesser than the denominator value. Then the fraction is denoted as proper fraction.

Definition of Improper fraction:

When the numerator  is greater than the denominator value. Then the fraction is denoted as Improper fraction.

Definition of compound fraction:

We can write the improper fraction is the combination of the whole number and then the proper fraction. This fraction is known as compound fraction.

Solving compound fraction problems

Addition of compound fractions:

Example 1:

Solving 2 (1)/(2) + 3 (1)/(2)

Solution:

First we can convert the compound fractions into improper fraction. It can be converted in the following manner

(((2*2)+1))/(2)+ (((3*2)+1))/(2)

That is (5)/(2)+(7)/(2)

Here the denominators are same so we can directly add the numerators otherwise we can take the L.C.D and then add the numerator values.

Adding this we can get, (12)/(2)

Divided by 2 we can get 6

This is the solution of the given fraction.

Subtraction of compound fraction:

Subtraction of compound fractions is similar to addition instead of + we can use the sign -.

Multiplication of compound fractions:

Example 2:

Solving 1 (1)/(3) x 2 (5)/(4)

Solution:

First we can convert the compound fractions into improper fraction. It can be converted in the following manner

(((1*3)+1))/(3) x (((2*4)+5))/(4)

that is (4)/(3) x (13)/(4)

Now we can multiply the numerator and denominator terms separately. Then we can simplify the fraction.

then we can get (13)/(3)

This is the solution of the given fraction.


Division of compound fractions:

Here we can convert the mixed numbers into improper fraction and then take the reciprocal of the second fraction. Now we have to multiply the fractions.

Linear Combination of Random Variables

The random variables have the countable values for their probability in the statistics. The sum of the probability for the random variable is one, then the random variable is said to be discrete random variable. The random variable is the measurable function. Here the mean, variance and standard deviation for the random variables are calculated. This article contains the information about the linear combination of random variables in the probability theory and statistics.

Formula used for linear combination of random variables:

The formulas used to determine the mean, variance and the standard deviation in the random variable are

Mean = `sum x p(x)`

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

In this above mentioned formula the `(sum x p(x)) ^2` refers the squared value of the mean in the statistics.

Standard deviation = `sqrt (variance)`

In the above mentioned formula x refers the given set of discrete random values and p(x) refers the probability value for the random variables in the statistics.

Examples for linear combination of random variables:

Example 1 to linear combination of random variables:

Predict the mean, variance and standard deviation for the random variables.

x	20	40	60	80	90
P(x)	0.11	0.12	0.34	0.24	0.19

Solution:

Mean = `sum x p(x)`  

Mean = 20 (0.11) +40 (0.12) +60 (0.34) +80 (0.24) + 90(0.19)

Mean = 2.2+ 4.8 + 20.4 + 19.2 + 17.1                   

Mean = 63.7

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

Variance = ((0.11) (20) (20) + (0.12) (40) (40) + (0.34) (60) (60) + (0.24) (80) (80) + (0.19) (90) (90)) - (63.7)² 

Variance = (44+ 192+ 1224 + 1536+ 1539) -4057.69

Variance = 4535 -4057.69

Variance = 477.31

Standard deviation = `sqrt (variance)`

Standard deviation = `sqrt (477.31)`

Standard deviation = 21.847

The mean value is 63.7, variance is 477.31 and the standard deviation is 21.847 for the given random variables.  

Algebra is widely used in day to day activities watch out for my forthcoming posts on Dividing Fraction and algebra course. I am sure they will be helpful.

Example 2 to linear combination of random variables:         

Predict the mean, variance and standard deviation of random variables.  

  

x	10	20	30	40	50
P(x)	0.20	0.12	0.22	0.13	0.33

Solution:

Mean = `sum x p(x)`

Mean = 10(0.20) +20(0.12) + 30(0.22) +40(0.13) + 50(0.33)

Mean = 2.0 +2.4 +6.6+ 5.2+ 16.5

Mean = 32.7

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

Variance = ((0.20) (10) (10) + (0.12) (20) (20) + (0.22) (30) (30) + (0.13) (40) (40) + (0.33) (50) (50)) - (32.7)² 

Variance = (20+48+ 132 + 208 +825) -1069.29

Variance = 1233 -1069.29                                     

Variance = 163.71

Standard deviation = `sqrt (variance)`

Standard deviation = `sqrt (163.71)`

Standard deviation = 12.79

The mean is 32.7, variance is 163.71 and the standard deviation is 12.79 for the given random variables.

Make Ellipse Equation

Under the picture that relates the foci, eccentricity, and length of the main and negligible axes. Derive the identities if we remember that if T is a point on the ellipse the addition of the distances from T to the two foci is a steady.  Derive the equation for the ellipse with vertical main axis.  The derivative for the horizontal main axis case is similar.

Vertical Ellipse and Horizondal Ellipse

Making ellipse equation:

Consider the upside of the ellipse.  The addition of the distances from the two foci to the up side of the make a ellipse is

(b –c) + (b + c) = 2b

The square of is given by

4b2

Then consider the right hand side of the ellipse.

See that the distances from each side to the right hand side is the equal.  The square of sum is

(2d)2 = 4d2

Using the Pythagorean Theorem,

d2 = a2 + c2

Replacement and setting these same to each other gives

4b2 = 4(a2 + c2)

So that

b2 = a2 + c2

Place the equation in the particularly simple form

`(x^2)/(a^2) + (y^2)/(b^2) =1 (or) (x^2)/(b^2) + (y^2)/(a^2) =1`

Make ellipse equations - Examples:

Make ellipse equations - Example1:

Given the following equation

9x2 + 4y2 = 36

a) Solve the x and y intercepts of the graph of the equation.

b) Solve the co-ordinates of the foci.

c) Solve the length of the main and slight axes.

d) Make the graph of the equation.

Solution:

9x2 / 36 + 4y2 / 36 = 1

x2 / 4 + y2 / 9 = 1

x2 / 22 + y2 / 32 = 1

a = 3 and b = 2 (NOTE: a >b) .

Set y = 0 in the equation get and Solve the xaxes

x2 / 22  = 1

Solve for x.

x2  = 22

x = ± 2

Set x = 0 in the equation get and Solve the y axes

y2 / 32 = 1

Solve for y.

y2  = 32

y = ± 3

b) We need to solve c first.

c2 = a2 - b2

a and b were found in part a).

c2 = 32 - 22

c2 = 5

Solve for c.

c = +- (5)1/2

The foci are    F1 (0, (5)1/2) and  F2 (0 , -(5)1/2)



c) The main axis length is given by  2 a = 6.

The slight axis length is given by  2 b = 4.

d) The graph of the equation is


Example Ellipse

Equivalence Properties of Equality Tutor

Tutor is nothing but the teacher or instructor give notes to the students who comes through online. Tutor teaches a certain topic to student what they want by online. Here tutor give instruction about the equivalence property of equality to students by online. Equality represents that both side of the equation is of equal status or not. To represent this equality; we use the equivalence property of equality.

Equivalence Properties of Equality Tutor:

There are three properties under the equivalence of equality. These properties are applicable for the arithmetic operation.

Reflexive property:

Reflexive property is said to be any number that is equal to same number.

Example: a = a

Symmetric property:

Symmetric property is said to a number x is equal to the number y; then the number y is equal to the number x.

x = y then y = x

Transitive property:

Transitive Property is said to be a number an equal to the number b then the number b equal to the number c; then we can say the number an equal to the number c.

a = b; b = c and then a = c

Example Problems – Equivalence Properties of Equality Tutor:

Example 1:

Verify the following statement is transitive Property

M + 3 = 7; 7 = 4 + 3 and then 4 + 3 = 7

Solution:

Given: M + 3 = 7; 7 = 4 + 3 and then 4 + 3 = 7

Let take M + 3 = 7 ---- > equation 1

7 = 4 + 3 ---- > equation 2

4 + 3 = 7 ----- > equation 3

Step 1: Find the value of M in equation 1.

M + 3 = 7

Step 2: Subtract the number 3 on each side of the equation 1, we get

M + 3 – 3 = 7 – 3

M = 4

Thus 4 + 3 = 7 --- equation 4

That the equation 1 and equation 4 are equal,

4 + 3 = 4 + 3

Thus we proved the given statement is transitive property of equivalence.

Example 2:

Which of the following is the symmetric property?

a) m = n and n = m

b) m = 1/m and m= m

c) m + 1 = m - 1

d) m = 1 + m

Solution:

Symmetric property: x = y; y = x

Here the option ‘’a’’ refers the symmetric property.

m = n and n = m

Answer: a

These are the example problem istruct by the tutor about the equivalence properties of equality.

Fraction in Containers

Fractions are method of writing numbers that are not  whole numbers. Some fractions are written as  `1/2` ,`16/41` ,`3/2`. Fractions contains of two numbers, numerator and denominator. In the general form,

Numerator
Fraction = ----------------
Denominator

Where, numerator is at  top and denominator on the bottom.

For example,

In the fraction `2/3` , 2 is numerator and 3 is denominator. Now we see the problems for solving a fractions.


Conditions for fractions:

Some of the conditions are there for fraction addition and subtraction is given as,

To add the fractions the denominator must be same, then the numerators can is counted directly as same as numbers.

If the fractions the denominator is not same means, make them same number by multiplying and dividing by same scale factor.

Fraction subtraction is similar to the fraction addition. Let us see some of example problems using containers with same denominator and various denominators in detail and the example problems for containers fractions for better understand.

Example Problems for fraction in containers:

Example 1:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of eight liter in which four liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `4/8`

The fractions are `1/2` and `4/8`

`4/8` can also be written as `1/2` (divide my 2 on both numerator and denominator.)

Total amount of water = `1/2 + 1/2`

= `(1+1)/2`

= `2/2`

= `1/1`

= 1

Example 2:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of eight liter in which six liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `6/8`

The fractions are `1/2 ` and `6/8`

Total amount of water =` 1/2 + 6/8`

= `1/2 * 4/4 + 6/8`

= `4/8 + 6/8`

= `(4+ 6)/8`

= `10/8`

= `5/4`


Example 3:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of three liter in which one liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `1/3`

The fractions are `1/2` and `1/3`

Total amount of water = `1/2 + 1/3`

= `1/2 * 3/3 + 1/3 * 2/2`

= `3/6 + 2/6`

= `5/6`

Sum of two Distributions

In sum of probability distributions and statistics, a probability will make the value of a random variable  or the probability of the value drawing  within a particular interval. The sum probability distribution defiines the measure of possible values like random variable can accomplish and the probability sum that the value of the random variable is within subset of that range.

Probability distributions:

Random Experiment and Sample Space

• Any experiment cannot be predicted in before, but is one of the set of possible outcomes, which is known as random experiment.
• An experiment performed frequently, every repetition is known as trial.
• The collection of all likely outcomes in random experiment is denoted the sample space.
• The sample space of a random experiment is denoted as a set S that includes all possible outcomes of the experiment
•  For simple experiments, the sample space may be exactly the collection  of possible outcomes.
• The sample space is a mathematically best set which contains the possible outcomes and perhaps other elements as well.
•  For example, if the testing is to draw a standard die and evidence the outcome, the sample space is S 1 2 3 4 5 6 , the set of possible outcomes.

Probability Distributions
      Let S acting as sample space with random experiment.
     Then X : x₁ x₂ x₃ …x_n
      P(X) is known as probability distribution of x.
Mean and variance of  Addition of two distributions:
Mean is E(X) = ∑ pi xi
variance is given below,
 

Examples:

1)Find the sum of mean of two distributions for the given below?

X	0	1	2
P(X)= x	0.25	.20	.25

X	0	1	2
P(X)= x	0.15	.20	.35

Solution:
Mean for table 1.
Mean E(X)1 = ∑ pi xi
            = 0*0.25 + 1*0.20+2*0.25
                       =0 + 0.20 +0.50  = 0.70
           Mean for table 2.
Mean E(X) 2 = ∑ pi xi
                    = 0*.15+1*0.20+2*0.35=0+.20+0.70
                 = 0.90
Sum of two distributions are E(X)1 + E(X)2= 0.70+0.90 = 1.6
2)Find out sum the variance of the following two distributions?

X	0	1	2
P(X)= x	0.25	.20	.25

X	0	1	2
P(X)= x	0.15	.20	.35

Congruent tTiangles Worksheet

In geometry, two figures are congruent if they have the same shape and size. More formally, 2 sets of points are called congruent if, and only if, one can be transformed into the other by an isometry, i.e., a combination of translations, rotations and reflections.

(Source-Wikipedia)

Congruent triangles worksheet -Example 1:

Let PQRS be a parallelogram and PR be one of its diagonals. What can you say about triangles PQRC andRSP?

Solution:
Step 1: In a four-sided figure, opposite sides are congruent. Hence sides

Step 2: QR and PS are congruent, and also sides PQ and RS are congruent.
Step 3: In a four-sided figure opposite angles are congruent.

Step 4: Hence angles PQR and RSP  are  four side figure.
Step 5: Two sides and an included angle of triangle PQR are congruent to two corresponding sides and an included angle in triangle RSP. According to the above postulate the two triangles PQR and RSP are congruent.

Congruent triangles worksheet-Example 2:

To solve the  triangle ABC congruent to triangle EFV by SAS?

Solution:
Step 1:Segment AB is congruent
Step 2: To segment of ST is
Step 3:  AB = EF = 4.
Step 4: Angle B is congruent to
Step 5: angle T because angle B = angle F = 100 degrees.
Step 6:Segment BC is congruent to segment FV because BC = FV = 5.
Step 7: ABC is congruent EFV is a SAS.



Congruent triangles worksheet-Example 3:
To solve the angle is QR is congruent to angle ST.

Solution:
Step 1: The triangle RPQ is congruent to triangle RTS.             
Step 2: Angle P is congruent to angle S because that information is given  in the diagram.
Step 3: Segment PR is congruent to segment RS because that information is given in the diagram.
Step 4: Angle QRP is congruent to angle SRT because vertical angles are congruent.
Step 5: Triangle SPQ is congruent to triangle RTS by Angle-Side-Angle.
Step 6: The triangles arecongruent, you know that all orresponding parts must be congruent.
Step 7:By CPCTC, segment BC is congruent to segment CE.